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Gaussian Integral

The Gaussian integral is the integral of the function eโˆ’x2e^{-x^2} over the entire real line, and it is given by:

โˆซโˆ’โˆžโˆžeโˆ’x2dx=ฯ€\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

First, we square the integral:

(โˆซโˆ’โˆžโˆžeโˆ’x2dx)2=โˆซโˆ’โˆžโˆžโˆซโˆ’โˆžโˆžeโˆ’x2eโˆ’y2dxdy\left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx dy

Next, we change to polar coordinates:

โˆซ0โˆžโˆซ02ฯ€eโˆ’r2rdฮธdr=ฯ€\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} r d\theta dr = \pi

To evaluate this integral, we use the substitution u=r2u = r^2 and du=2rdrdu = 2r dr, which gives:

12โˆซ0โˆžeโˆ’udu=12\frac{1}{2}\int_{0}^{\infty} e^{-u} du = \frac{1}{2}

Therefore, we have:

(โˆซโˆ’โˆžโˆžeโˆ’x2dx)2=ฯ€โ‡’โˆซโˆ’โˆžโˆžeโˆ’x2dx=ฯ€\left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)^2 = \pi \quad \Rightarrow \quad \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

And this completes the proof.